In the given arrangement,the spring constant | vedclass

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(C) In this system,both blocks $M$ and $m$ move together as a single unit because the friction between them is sufficient to prevent relative motion f

Vedclass · Accepted Answer

$2\sqrt{2}\pi$

Vedclass · Answer

(C) In this system,both blocks $M$ and $m$ move together as a single unit because the friction between them is sufficient to prevent relative motion for small oscillations.Since the surface is smooth,the total mass of the oscillating system is $M_{total} = M + m = 3\,kg + 1\,kg = 4\,kg$.The spring constant is $k = 2\,N\,m^{-1}$.The time period $T$ of a spring-mass system is given by the formula $T = 2\pi \sqrt{\frac{M_{total}}{k}}$.Substituting the values,we get $T = 2\pi \sqrt{\frac{4}{2}} = 2\pi \sqrt{2}$.Thus,the time period of the $SHM$ executed by the system is $2\sqrt{2}\pi\,s$.

In the given arrangement,the spring constant $k$ has a value of $2\,N\,m^{-1}$,mass $M = 3\,kg$,and mass $m = 1\,kg$. Mass $M$ is in contact with a smooth surface. The coefficient of friction between the two blocks is $0.1$. The time period of $SHM$ executed by the system is

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